What is the extreme f(x,y)=(x-8)^2+(y-5)^2+16 ?

The extreme f(x,y)=(x-8)^2+(y-5)^2+16 is Minimum(8,5)

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extreme f(x,y)=(x-8)^2+(y-5)^2+16

extreme points

f (x, y) = (x - 8)^{2} + (y - 5)^{2} + 16

Minimum (8, 5)

Solution steps

Find the critical points:

(8, 5)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(8, 5) :

Minimum

Minimum (8, 5)

f (x) = x e^{3 x^{2}}

f (x, y) = 2 x^{2} + y^{4} - 16 x y

f (x, y) = (7 x + 9 y + 8) e^{(- (x^{2} + y^{2}))}

x^{3} - 3 x + 3 x y^{2}

f (x) = \frac{x}{( x + 3 ) ^{2}}

What is the extreme f(x,y)=(x-8)^2+(y-5)^2+16 ?
The extreme f(x,y)=(x-8)^2+(y-5)^2+16 is Minimum(8,5)